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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p><dfn class="terminology">Example 1</dfn> Find the general solution of the following equations</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_8.html ./knowl/eq7_8.html">
\begin{equation*}
{\bf x}^{\prime}={\bf A}\, {\bf x}=\left(\begin{array}{cc}
-2 &amp; 1 \\
1 &amp; -2\\
\end{array}
\right) {\bf x}.
\end{equation*}
</div>
<p class="continuation"><dfn class="terminology">Solution</dfn> Consider</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_8.html ./knowl/eq7_8.html">
\begin{equation}
({\bf A}-r {\bf I}) \vec{\xi}={\bf 0}.\tag{6.2.2}
\end{equation}
</div>
<p class="continuation">For non-zero solutions of <span class="process-math">\(\vec{\xi}\text{,}\)</span> we require that</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_8.html ./knowl/eq7_8.html">
\begin{equation*}
|{\bf A}-r {\bf I}|=0 \rightarrow
\left| \begin{array}{cc}
-2-r &amp; 1\\
1 &amp; -2-r
\end{array}
\right|=0 \rightarrow r^2+4r+3=0 \rightarrow r_1=-1, r_2=-3.
\end{equation*}
</div>
<p class="continuation">For <span class="process-math">\(r_1=-1\text{,}\)</span> from (<a href="" class="xref" data-knowl="./knowl/eq7_8.html" title="Equation 6.2.2">(6.2.2)</a>):</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_8.html ./knowl/eq7_8.html">
\begin{equation*}
\left(
\begin{array}{cc}
-2-r_1 &amp; 1\\
1 &amp; -2-r_1
\end{array}
\right) \vec{\xi}^{(1)}={\bf 0} \rightarrow 
\left(
\begin{array}{cc}
-1 &amp; 1\\
1 &amp; -1
\end{array}
\right) 
\left(
\begin{array}{c}
\xi^{(1)}_1\\
\xi^{(1)}_2
\end{array}
\right)={\bf 0} \rightarrow
\xi^{(1)}_1=\xi^{(1)}_2.
\end{equation*}
</div>
<p class="continuation">Choose that <span class="process-math">\(\xi^{(1)}_1=\xi^{(1)}_2=1\text{,}\)</span> then</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_8.html ./knowl/eq7_8.html">
\begin{equation*}
\vec{\xi}^{(1)}=\left(
\begin{array}{c}
1\\
1
\end{array}
\right),
\end{equation*}
</div>
<p class="continuation">and</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_8.html ./knowl/eq7_8.html">
\begin{equation*}
{\bf x}^{(1)}=\left(
\begin{array}{c}
1\\
1
\end{array}
\right) e^{-t}.
\end{equation*}
</div>
<p class="continuation">For <span class="process-math">\(r_2=-3\text{,}\)</span> from (<a href="" class="xref" data-knowl="./knowl/eq7_8.html" title="Equation 6.2.2">(6.2.2)</a>), we have</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_8.html ./knowl/eq7_8.html">
\begin{equation*}
\left(
\begin{array}{cc}
-2-r_2 &amp; 1\\
1 &amp; -2-r_2
\end{array}
\right) \vec{\xi}^{(2)}=\left(
\begin{array}{cc}
1 &amp; 1\\
1 &amp; 1
\end{array}
\right) \left(
\begin{array}{c}
\xi^{(2)}_1\\
\xi^{(2)}_2
\end{array}
\right)=0 \rightarrow \xi^{(2)}_1=-\xi^{(2)}_2.
\end{equation*}
</div>
<p class="continuation">Choose <span class="process-math">\(\xi^{(2)}_1=1\text{,}\)</span> we have</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_8.html ./knowl/eq7_8.html">
\begin{equation*}
\vec{\xi}^{(2)}=\left(\begin{array}{c}
1\\
-1
\end{array}\right)
\end{equation*}
</div>
<p class="continuation">and</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_8.html ./knowl/eq7_8.html">
\begin{equation*}
{\bf x}^{(2)}=\left(\begin{array}{c}
1\\
-1
\end{array}\right) e^{-3t}.
\end{equation*}
</div>
<span class="incontext"><a href="sec6_2.html#p-260" class="internal">in-context</a></span>
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